高考理科数学一轮复习考点大题专项训练高考大题专项练1

高考大题专项练1 高考中的函数与导数

1.(2015山西四校联考)已知f (x )=ln x-x+a+1.

(1)若存在x ∈(0,+∞)使得f (x )≥0成立,求a 的取值范围; (2)求证:当x>1时,在(1)的条件下,1

2x 2+ax-a>x ln x+1

2成立. 解:(1)原题即为存在x>0使得ln x-x+a+1≥0,

∴a ≥-ln x+x-1,令g (x )=-ln x+x-1,

则g'(x )=-1

x +1=x -1

x .令g'(x )=0,解得x=1.

∵当0

当x>1时,g'(x )>0,g (x )为增函数,

∴g (x )min =g (1)=0,a ≥g (1)=0.

故a 的取值范围是[0,+∞).

(2)证明:原不等式可化为1

2x 2+ax-x ln x-a-1

2>0(x>1,a ≥0). 令G (x )=1

2x 2+ax-x ln x-a-12,则G (1)=0. 由(1)可知x-ln x-1>0,

则G'(x )=x+a-ln x-1≥x-ln x-1>0,

∴G (x )在(1,+∞)上单调递增, ∴G (x )>G (1)=0成立, ∴1

2x 2+ax-x ln x-a-1

2>0成立,

即12

x 2+ax-a>x 1n x+12

成立.

〚导学号92950922〛

2.(2015河南新乡调研)已知函数f (x )=x-(a+1)ln x-a

x

(a ∈R ),g (x )=12

x 2+e x -x e x . (1)当x ∈[1,e]时,求f (x )的最小值;

(2)当a<1时,若存在x 1∈[e,e 2],使得对任意的x 2∈[-2,0],f (x 1)

(x -1)(x -a )

x 2

. ①当a ≤1时,x ∈[1,e],f'(x )≥0,

f (x )为增函数,f (x )min =f (1)=1-a.

②当1

x ∈[a ,e]时,f'(x )≥0,f (x )为增函数. 所以f (x )min =f (a )=a-(a+1)ln a-1.

③当a ≥e 时,x ∈[1,e]时,f'(x )≤0,

f (x )在[1,e]上为减函数. f (x )min =f (e)=e -(a+1)-a e

. 综上,当a ≤1时,f (x )min =1-a ; 当1

.

(2)由题意知:f (x )(x ∈[e,e 2])的最小值小于g (x )(x ∈[-2,0])的最小值. 由(1)知f (x )在[e,e 2]上单调递增,f (x )min =f (e)=e -(a+1)-a e

. g'(x )=(1-e x )x.

当x ∈[-2,0]时g'(x )≤0,g (x )为减函数,g (x )min =g (0)=1, 所以

e -(a+1)-a

e <1,即

a>e 2-2e e+1,

所以a 的取值范围为(e 2-2e

e+1

,1).

〚导学号92950923〛

3.(2015河北唐山二模)已知f (x )=x+1

x +a ln x ,其中a ∈R . (1)设f (x )的极小值点为x=t ,请将a 用t 表示; (2)记f (x )的极小值为g (t ),证明:

①g (t )=g (1

t );

②函数y=g (t )恰有两个零点,且互为倒数.

解:(1)f'(x )=1-1

x 2+a

x =

x 2+ax -1x 2,t=-a+√a 2+4

2

>0, 当x ∈(0,t )时,f ¢(x )<0,f (x )单调递减; 当x ∈(t ,+∞)时,f ¢(x )>0,f (x )单调递增. 由f ¢(t )=0得a=1t

-t.

(2)①由(1)知f (x )的极小值为g (t )=t+1

t

+(1t

-t)ln t , 则g (1

t )=1

t +t+(t -1

t )ln 1

t =t+1

t +(1

t

-t)ln t=g (t ).

②g ¢(t )=-(1+1t

2)ln t ,

当t ∈(0,1)时,g ¢(t )>0,g (t )单调递增; 当t ∈(1,+∞)时,g ¢(t )<0,g (t )单调递减. 又g (1

e 2)=g (e 2)=3

e 2-e 2<0,g (1)=2>0, 分别存在唯一的c ∈(1

e 2,1)和d ∈(1,e 2),

使得g (c )=g (d )=0,且cd=1, 所以y=g (t )有两个零点且互为倒数.

〚导学号92950924〛

4.(2015河北保定高三调研)已知函数f (x )=ln x+ax-a 2x 2(a ≥0). (1)若x=1是函数y=f (x )的极值点,求a 的值;

(2)若f (x )<0在定义域内恒成立,求实数a 的取值范围. 解:(1)函数的定义域为(0,+∞),

f'(x )=

-2a 2x 2+ax+1

x

. 因为x=1是函数y=f (x )的极值点, 所以f'(1)=1+a-2a 2=0,

解得a=-1

2

或a=1.又a ≥0,所以a=-12

(舍去).

经检验当a=1时,x=1是函数y=f (x )的极值点,所以a=1.

(2)当a=0时,f (x )=ln x ,显然在定义域内不满足f (x )<0恒成立;当a>0时,令f'(x )=(2ax+1)(-ax+1)

x

=0,得x 1=-1

2a (舍去),x 2=1

a ,

所以f'(x ),f (x )的变化情况如下表:

所以f (x )max =f (1

a )=ln 1

a <0,

∴a>1.

综上可得实数a 的取值范围是(1,+∞). 〚导学号92950925〛

5.(2015课标全国Ⅱ,理21)设函数f (x )=e mx +x 2-mx. (1)证明:f (x )在(-∞,0)上单调递减,在(0,+∞)上单调递增;

(2)若对于任意x 1,x 2∈[-1,1],都有|f (x 1)-f (x 2)|≤e -1,求m 的取值范围. 解:(1)f'(x )=m (e mx -1)+2x.

若m ≥0,则当x ∈(-∞,0)时,e mx -1≤0,f'(x )<0; 当x ∈(0,+∞)时,e mx -1≥0,f'(x )>0. 若m<0,则当x ∈(-∞,0)时,e mx -1>0,f'(x )<0; 当x ∈(0,+∞)时,e mx -1<0,f'(x )>0.

所以,f (x )在(-∞,0)上单调递减,在(0,+∞)上单调递增.